Net Ionic Equations and Solubility

Net Ionic Equations and Solubility

Assessment

Interactive Video

Chemistry

9th - 10th Grade

Practice Problem

Hard

Created by

Mia Campbell

FREE Resource

The video tutorial explains how to write a balanced net ionic equation for the reaction between lead(II) nitrate and sodium fluoride. It begins with balancing the molecular equation, identifying the solubility of compounds, and determining the precipitate. The tutorial then demonstrates how to form the complete ionic equation by splitting strong electrolytes into ions, followed by removing spectator ions to derive the net ionic equation. The video concludes by ensuring the net charge and atom balance in the final equation.

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10 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the first step in writing a balanced net ionic equation?

Balance the molecular equation

Identify the spectator ions

Determine the solubility of compounds

Write the complete ionic equation

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How do you balance the molecular equation for the reaction between lead(II) nitrate and sodium fluoride?

By adding more reactants

By adjusting the coefficients

By changing the products

By removing spectator ions

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which compound is identified as the precipitate in the reaction?

Sodium nitrate

Sodium fluoride

Lead(II) nitrate

Lead(II) fluoride

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the solubility status of lead(II) fluoride in the reaction?

Highly soluble

Completely soluble

Insoluble

Slightly soluble

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the purpose of writing the complete ionic equation?

To balance the charges

To identify the products

To split strong electrolytes into ions

To determine the solubility

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which ions are considered spectator ions in this reaction?

All ions

Lead ions

Fluoride ions

Nitrate and sodium ions

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the net ionic equation derived from the reaction?

Pb(NO3)2 + 2NaF → PbF2 + 2NaNO3

Na^+ + NO3^- → NaNO3

2Na^+ + 2NO3^- → 2NaNO3

Pb^2+ + 2F^- → PbF2

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